Question: Find the distance between the point ${(1, -8)}$ and the line $\enspace {y = x + 3}\thinspace$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
First, find the equation of the perpendicular line that passes through ${(1, -8)}$ The slope of the blue line is ${1}$ , and its negative reciprocal is ${-1}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -x + b}\thinspace$ We can plug our point, ${(1, -8)}$ , into this equation to solve for ${b}$ , the y-intercept. $-8 = {-}(1) + {b}$ $-8 = -1 + {b}$ $-8 + 1 = {b} = -7$ The equation of the perpendicular line is $\enspace {y = -x - 7}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(-5, -2)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(1, -8)}$ and ${(-5, -2)}$ gives us: $\sqrt{( {1} - {-5} )^2 + ( {-8} - {-2} )^2}$ $= \sqrt{( 6 )^2 + ( -6 )^2} = \sqrt{72} = 6\sqrt{2}$ The distance between the point ${(1, -8)}$ and the line $\thinspace {y = x + 3}\enspace$ is $\thinspace6\sqrt{2}$.